MCQ
$\int_{}^{} {\frac{{dx}}{{{e^x} - 1}} = } $
  • $\ln (1 - {e^{ - x}}) + c$
  • B
    $ - \ln (1 - {e^{ - x}}) + c$
  • C
    $\ln ({e^x} - 1) + c$
  • D
    None of these

Answer

Correct option: A.
$\ln (1 - {e^{ - x}}) + c$
a
(a)$\int_{}^{} {\frac{{dx}}{{{e^x} - 1}} = \int_{}^{} {\frac{{{e^{ - x}}}}{{1 - {e^{ - x}}}}\,dx} } $
Put $1 - {e^{ - x}} = t \Rightarrow {e^{ - x}}dx = dt,$ then it reduces to
$\int_{}^{} {\frac{{dt}}{t} = \log t + c} = \log (1 - {e^{ - x}}) + c.$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Let $A=\left(\begin{array}{cc}1 & 2 \\ -2 & -5\end{array}\right)$. Let $\alpha, \beta \in R$ be such that $\alpha A^{2}+\beta A=2 I$. Then $\alpha+\beta$ is equal to -
The differential equation whose solution is $y = {c_1}\cos ax + {c_2}\sin ax$ is(Where ${c_1},\;{c_2}$are arbitrary constants)
If $x = a\left( {t - {1 \over t}} \right)\,,y = a$ $\left( {t + {1 \over t}} \right)$ then ${{dy} \over {dx}} = $
If $f(x) \, \& \,g(x)$ are inverse functions of each other such that $f(1) = 3\, \& \,f(3) = 1,$ then $\int\limits_1^3 {\left( {g(x) + \frac{x}{{f'\left( {g\left( x \right)} \right)}}} \right)} dx$ is equal to -
If $|\vec{a}-\vec{b}|=|\vec{a}+\vec{b}|$ then the angle between $\vec{a}$ and $\vec{b}$ will be $-$
Choose the correct answer from the given four options.If two events are independent, then :
The value of $\int\limits_0^1 {\left( {\prod\limits_{r = 1}^n {(x + r)} } \right)\left( {\sum\limits_{k = 1}^n {\frac{1}{{x + k}}} } \right)} dx$ equals
If the projection of the vector $\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ on the sum of the two vectors $2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}$ and $-\lambda \hat{i}+2 \hat{j}+3 \hat{k}$ is $1,$ then $\lambda$ is equal to ..... .
Let $\quad P_1=I=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right], \quad P_2=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right], \quad P_3=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right], \quad P_4=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right], \quad P_5=\left[\begin{array}{lll}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{array}\right]$,

$P_6=\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]$ and $X=\sum_{k=1}^6 P_k\left[\begin{array}{lll}2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1\end{array}\right] P_k^{\top}$

where $P _{ K }^{ T }$ denotes the transpose of the matrix $P _{ K }$. Then which of the following options is/are correct?

$(1)$ $X -30 I$ is an invertible matrix

$(2)$ The sum of diagonal entries of $X$ is 18

$(3)$ If $X \left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\alpha\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$, then $\alpha=30$

$(4)$ $X$ is a symmetric matrix

Let ${I_1} = \int\limits_0^1 {\frac{{{e^x}}}{{\left( {1 + x} \right)}}} \,dx$ and ${I_2} = \int\limits_0^1 {\frac{{{x^2}}}{{{e^{{x^3}}}\left( {2 - {x^3}} \right)}}} \,dx$ then value of $\frac{{{I_1}}}{{{I_2}}}$ is equal to